Liouville-Arnold integrability of the pentagram map on closed polygons Valentin Ovsienko Richard Evan Schwartz Serge Tabachnikov Abstract The pentagram map is a discrete dynamical system defined on the moduli space of poly- gons in the projective plane. This map has recently attracted a considerable interest, mostly because its connection to a number of different domains, such as: classical projective geom- etry, algebraic combinatorics, moduli spaces, cluster algebras and integrable systems. Integrability of the pentagram map was conjectured in [19] and proved in [15] for a larger space of twisted polygons. In this paper, we prove the initial conjecture that the pentagram map is completely integrable on the moduli space of closed polygons. In the case of convex polygons in the real projective plane, this result implies the existence of a toric foliation on the moduli space. The leaves of the foliation carry affine structure and the dynamics of the pentagram map is quasi-periodic. Our proof is based on an invariant Poisson structure on the space of twisted polygons. We prove that the Hamiltonian vector fields corresponding to the monodoromy invariants preserve the space of closed polygons and define an invariant affine structure on the level surfaces of the monodromy invariants. Contents 1 Introduction 2 1.1 Integrability problem and known results ....................... 3 1.2 The main theorem ................................... 4 1.3 Related topics ...................................... 5 2 Integrability on the space of twisted n-gons 6 2.1 The space P n ...................................... 6 2.2 The corner coordinates ................................. 7 2.3 Rescaling and the spectral parameter ......................... 8 2.4 The Poisson bracket .................................. 9 2.5 The rank of the Poisson bracket and the Casimir functions ............. 11 2.6 Two constructions of the monodromy invariants ................... 12 2.7 The monodromy invariants Poisson commute .................... 14
Transcript
Liouville-Arnold integrability of the pentagram map on closed
polygons
Valentin Ovsienko Richard Evan Schwartz Serge Tabachnikov
Abstract
The pentagram map is a discrete dynamical system defined on the moduli space of poly-gons in the projective plane. This map has recently attracted a considerable interest, mostlybecause its connection to a number of different domains, such as: classical projective geom-etry, algebraic combinatorics, moduli spaces, cluster algebras and integrable systems.
Integrability of the pentagram map was conjectured in [19] and proved in [15] for a largerspace of twisted polygons. In this paper, we prove the initial conjecture that the pentagrammap is completely integrable on the moduli space of closed polygons. In the case of convexpolygons in the real projective plane, this result implies the existence of a toric foliation onthe moduli space. The leaves of the foliation carry affine structure and the dynamics of thepentagram map is quasi-periodic. Our proof is based on an invariant Poisson structure onthe space of twisted polygons. We prove that the Hamiltonian vector fields correspondingto the monodoromy invariants preserve the space of closed polygons and define an invariantaffine structure on the level surfaces of the monodromy invariants.
The pentagram map is a geometric construction which carries one polygon to another. Given ann-gon P , the vertices of the image T (P ) under the pentagram map are the intersection pointsof consecutive combinatorially shortest diagonals of P . The left side of Figure 1 shows the basicconstruction. The right hand side shows the second iterate of the pentagram map. The seconditerate has the virtue that it acts in a canonical way on a labeled polygon, as indicated. Thefirst iterate also acts on labeled polygons, but one must make a choice of labeling scheme; seeSection 2.2. The simplest example of the pentagram map for pentagons was considered in [13].In the case of arbitrary n, the map was introduced in [17] and further studied in [18, 19].
The pentagram map is defined on any polygon whose points are in general position, andalso on some polygons whose points are not in general position. One sufficient condition forthe pentagram map to be well defined is that every consecutive triple of points is not collinear.However, this last condition is not invariant under the pentagram map.
The pentagram map commutes with projective transformations and thereby induces a (gener-ically defined) map
T : Cn → Cn (1.1)
where Cn is the moduli space of projective equivalence classes of n-gons in the projective plane.Mainly we are interested in the subspace C0n of projective classes of convex n-gons. The penta-gram map is entirely defined on C0n and preserves this subspace.
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P
T (P)2
P
T(P)
1
4
5
1
2
34
5
6
7
6
3
2
7
Figure 1: The pentagram map and its second iterate defined on a convex 7-gon
Note that the pentagram map can be defined over an arbitrary field. Usually, we restrictour considerations to the geometrically natural real case of convex n-gons in RP2. However, thecomplex case represents a special interest since the moduli space of n-gons in CP2 is a higheranalog of the moduli spaceM0,n (the moduli space of stable curves of genus zero with n distinctmarked points). Unless specified, we will be using the general notation P2 for the projectiveplane and PGL3 for the group of projective transformations.
1.1 Integrability problem and known results
The maps T : C5 → C5 and T : C6 → C6 are periodic. Indeed, there are maps T ′ : C5 → C5 andT ′ : C6 → C6, which differ from T only by composition with a cyclic relabelling, so that T ′ isthe identity on C5 and an involution on C6. See [17]. (These alternate labeling schemes are onlyconvenient for the cases n = 5, 6 so we do not use them below.)
The conjecture that the map (1.1) is completely integrable was formulated roughly in [17]and then more precisely in [19]. This conjecture was inspired by computer experiments in thecase n = 7. Figure 2 presents (a two-dimensional projection of) an orbit of a convex heptagonin RP2. (In the example, the bilateral symmetry of the initial heptagon causes the orbit to have2-dimensional closure, rather than a 3-dimensional closure, as one would expect from our mainresult below.)
The first results regarding the integrability of the pentagram map were proved for the pen-tagram map defined on a larger space, Pn, of twisted n-gons. A series of T -invariant functions(or first integrals) called the monodromy invariants, was constructed in [19]. In [15] (see also[14] for a short version), the complete integrability of T on Pn was proved with the help of aT -invariant Poisson structure, such that the monodromy invariants Poisson-commute.
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Figure 2: An orbit of the pentagram map on a heptagon
In [22], F. Soloviev found a Lax representation of the pentagram map and proved its algebraic-geometric integrability. The space of polygons (either Pn or Cn) is parametrized in terms of aspectral curve with marked points and a divisor. The spectral curve is determined by themonodromy invariants, and the divisor corresponds to a point on a torus – the Jacobi varietyof the spectral curve. These results allow one to construct explicit solutions formulas usingRiemann theta functions (i.e., the variables that determine the polygon as explicit functionsof time). Soloviev also deduces the invariant Poisson bracket of [15] from the Krichever-Phonguniversal formula.
Our result below has the same dynamical implications as that of Soloviev, in the case ofreal convex polygons. Soloviev’s approach is by way of algebraic integrability, and it has theadvantage that it identifies the invariant tori explicitly as certain Jacobi varieties. Our proof isin the framework of Liouville-Arnold integrability, and it is more direct and self-contained.
1.2 The main theorem
The main result of the present paper is to give a purely geometric proof of the following result.
Theorem 1. Almost every point of Cn lies on a T -invariant algebraic submanifold of dimension
d =
n− 4, n is oddn− 5, n is even.
(1.2)
that has a T -invariant affine structure.
Recall that an affine structure on a d-dimensional manifold is defined by a locally free actionof the d-dimensional Abelian Lie algebra, that is, by d commuting vector fields linearly inde-
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pendent at every point. (The vector fields, but not the canonical affine structure, depend on achoice of a basis for the Lie algebra.)
In the case of convex n-gons in the real projective plane, thanks to the compactness of thespace established in [18], our result reads:
Corollary 1.1. Almost every orbit in C0n lies on a finite union of smooth d-dimensional tori,where d is as in equation (1.2). The union of these tori has a T -invariant affine structure.
Hence, the orbit of almost every convex n-gon undergoes quasi-periodic motion under the penta-gram map. The above statement is closely related to the integrability theorem in the Liouville–Arnold sense [1].
Let us also mention that the dimension of the invariant sets given by (1.2) is precisely a halfof the dimension of Cn, provided n is odd, which is a usual, generic, situation for an integrablesystem. If n is even, then d = 1
2 dim Cn − 1 so that one can talk of “hyper-integrability”. Also,we remark that the cases n = 5, 6 are special in that the maps are periodic and hence the orbitsare just finite sets of points.
Our approach is based on the results of [19] and [15]. We prove that the level sets of the mon-odromy invariants on the subspace Cn ⊂ Pn are algebraic subvarieties of Cn of dimension (1.2).We then prove that the Hamiltonian vector fields corresponding to the invariant functions aretangent to Cn (and therefore to the level sets). Finally, we prove that the Hamiltonian vectorfields define an affine structure on a generic level set. The main calculation, which establishesthe needed independence of the monodromy invariants and their Hamiltonian vector fields, usesa trick that is similar in spirit to tropical algebra.
One point that is worth emphasizing is that our proof does not actually produce a symplectic(or Poisson) structure on the space Cn. Rather, we use the Poisson structure on the ambientspace Pn, together with the invariants, to produce enough commuting flows on Cn in order tofill out the level sets.
1.3 Related topics
The pentagram map is a particular example of a discrete integrable system. The main motiva-tion for studying this map is its relations to different subjects, such as: a) projective differentialgeometry; b) classical integrable systems and symplectic geometry; c) cluster algebras; d) alge-braic combinatorics of Coxeter frieze patterns. All these relations may be beneficial not onlyfor the study of the pentagram map, but also for the above mentioned subjects. Let us mentionhere some recent developments involving the pentagram map.
• The relation of T to the classical Boussinesq equation was essential for [15]. In particular,the Poisson bracket was obtained as a discretization of the (first) Adler-Gelfand-Dickeybracket related to the Boussinesq equation. We refer to [23, 24] and references therein formore information about different versions of the discrete Boussinesq equation.
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• In [20], surprising results of elementary projective geometry are obtained in terms of thepentagram map, its iterations and generalizations.
• In [21], special relations amongst the monodromy invariants are established for polygonsthat are inscribed into a conic.
• In [2], the pentagram map is related to Lie-Poisson loop groups and to dimer models.
• The paper [11] concerns discretizations of Adler-Gelfand-Dickey flows as multi-dimensionalgeneralizations of the pentagram map.
• A particularly interesting feature of the pentagram map is its relation to the theory ofcluster algebras developed by Fomin and Zelevinsky, see [3]. This relation was noticedin [15] and developed in [7], where the pentagram map on the space of twisted n-gonsis interpreted as a sequence of cluster algebra mutations, and an explicit formula for theiterations of T is calculated1.
• Extending Glick’s approach and developing the connection with cluster algebras, [6] intro-duces higher pentagram maps and proves their complete integrability using the machineryof weighted directed networks on surfaces.
• The structure of cluster manifold on the space Cn and the related notion of 2-frieze patternare investigated in [12].
• The singularities of the pentagram map are studied in [8]. A typical singularity disappearsafter a finite number of iterations (a confinement phenomenon).
• A version of higher-dimensional pentagram map is introduced and studied in [9].
2 Integrability on the space of twisted n-gons
In this section, we explain the proof of the main result in our paper [15], the Liouville-Arnoldintegrability of the pentagram map on the space of twisted n-gons. While we omit some technicaldetails, we take the opportunity to fill a gap in [15]: there we claimed that the monodromyinvariants Poisson commute, but our proof there had a flaw. Here we present a correct proof ofthis fact.
2.1 The space PnWe recall the definition of the space of twisted n-gons.
1This can be understood as a version of integrability or “complete solvability”.
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A twisted n-gon is a map φ : Z→ P2 such that
vi+n = M vi, (2.1)
for all i ∈ Z and some fixed element M ∈ PGL3 called the monodromy. We denote by Pnthe space of twisted n-gons modulo projective equivalence. The pentagram map extends to agenerically defined map T : Pn → Pn. The same geometric definition given for ordinary polygonsworks here (generically), and the construction commutes with projective transformations.
In the next section we will describe coordinates on Pn. These coordinates identify Pn asan open dense subset of R2n. Sometimes we will simply identify Pn with R2n. The space Cn ismuch more complicated; it is an open dense subset of a codimension 8 subvariety of R2n.
Remark 2.1. If n 6= 3m, then it seems useful to impose the simple condition that vi, vi+1, vi+2
are in general position for all i. With this condition, Pn is isomorphic to the space of differenceequations of the form
Vi = ai Vi−1 − bi Vi−2 + Vi−3, (2.2)
where ai, bi ∈ C or R are n-periodic: ai+n = ai and bi+n = bi, for all i. Therefore, Pn is just a2n-dimensional vector space, provided n 6= 3m. Let us also mention that the spectral theory ofdifference operators of type (2.2) is a classical domain (see [10] and references therein).
2.2 The corner coordinates
Following [19], we define local coordinates (x1, . . . , x2n) on the space Pn and give the explicitformula for the pentagram map.
Recall that the (inverse) cross ratio of 4 collinear points in P2 is given by
In this formulation, the line containing the points is identified with R ∪ ∞ by a projectivetransformation so that ti ∈ R for all i. Any identification yields the same final answer.
where (v, w) stands for the line through v, w ∈ P2, see Figure 3. The functions (x1, . . . , x2n)are cyclically ordered: xi+2n = xi. They provide a system of local coordinates on the space Pncalled the corner invariants, cf. [19].
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i
i+2
vi+1v
i−1
vi−2
v
v
Figure 3: Definition of the corner invariants
Remark 2.2. a) The right hand side of the second equation is obtained from the right handside of the first equation just by swapping the roles played by (+) and (−). In light of thisfact, it might seem more natural to label the variables so that the second equation defines x2i+1
rather than x2i+0. The corner invariants would then be indexed by odd integers. In Section 5we will present an alternate labelling scheme which makes the indices work out better.
b) Continuing in the same vein, we remark that there are two useful ways to label thecorner invariants. In [19] one uses the variables x1, x2, x3, x4, ... whereas in [15, 21] one usesthe variables x1, y1, x2, y2, .... The explicit correspondence between the two labeling schemesis x2i−1 → xi, x2i → yi. We call the former convention the flag convention whereas we callthe latter convention the vertex convention. The reason for the names is that the variablesx1, x2, x3, x4 naturally correspond to the flags of a polygon, as we will see in Section 5. Thevariables xi, yi correspond to the two flags incident to the ith vertex.
Let us give an explicit formula for the pentagram map in the corner coordinates. Follow-ing [15], we will choose the right labelling2 of the vertices of T (P ), see Figure 4. One then has(see [19]):
T ∗x2i−1 = x2i−11− x2i−3 x2i−21− x2i+1 x2i+2
, T ∗x2i = x2i+21− x2i+3 x2i+4
1− x2i−1 x2i, (2.5)
where T ∗xi stands for the pull-back of the coordinate functions.
2.3 Rescaling and the spectral parameter
Equation (2.5) has an immediate consequence: a scaling symmetry of the pentagram map.
2To avoid this choice between the left or right labelling one can consider the square T 2 of the pentagram map..
8
5
2
3
4
5
4
1
right
22
3
4
5
34
1
left
3
Figure 4: Left and right labelling
Consider a one-parameter group R∗ (or C∗ in the complex case) acting on the space Pnmultiplying the coordinates by s or s−1 according to parity:
Rt : (x1, x2, x3 . . . , x2n)→ (s x1, s−1 x2, s x3, . . . , s
−1x2n). (2.6)
It follows from (2.5), that the pentagram map commutes with the rescaling operation.We will call the parameter s of the rescaling symmetry the spectral parameter since it defines
a one-parameter deformation of the monodromy, Ms. Note that the notion of spectral parameteris extremely useful in the theory of integrable systems.
2.4 The Poisson bracket
Recall that a Poisson bracket on a manifold is a Lie bracket ., . on the space of functionssatisfying the Leibniz rule:
F,GH = F,GH +GF,H,
for all functions F,G and H. The Poisson bracket is an essential ingredient of the Liouville-Arnold integrability [1].
Define the following Poisson structure on Pn. For the coordinate functions we set
xi, xi+2 = (−1)i xi xi+2, (2.7)
and all other brackets vanish. In other words, the Poisson bracket xi, xj of two coordinatefunctions is different from zero if and only if |i − j| = 2. The Leibniz rule then allows one toextend the Poisson bracket to all polynomial (and rational) functions. Note that the Jacobiidentity obviously holds. Indeed, the bracket (2.7) has constant coefficients when considered inthe logarithmic coordinates log xi.
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Action on Monomials: Here is a more explicit description of how the bracket acts on mono-mials. Given monomials A and B, we form a bipartite graph, where the top vertices and thebottom vertices are both indexed by the set 1, ..., 2n. We join the top vertices ai to the bottomvertices bi±2 iff xi appears in A and xi±2 appears in B. Indices are taken cyclically, as usual. Welabel the edge joining ai to bi±2 with (±) if i is even and with (∓) if i is odd. Then A,B/AB isthe number of (+) signs minus the number of (−) signs. One derives this description by induc-tion and Leibniz’s rule. Figure 5 illustrates this for n = 6 and A = x3x4x5x9 and B = x1x5x6x7.In this case A,B/AB = 1. The thick lines are labelled with (−) and thin ones with (+).
101 2 3 4 6 95 7 8
1 2 3 4 6 95 7 8 10 11 12
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Figure 5: Graphical calculation of x3x4x5x9, x1x5x6x7
Alternatively, one may orient the graph according to the sign of the bracket between therespective variables: the thin edges are oriented downward and the thick one upward. Thecoefficient A,B/AB is the intersection number of this oriented graph with the horizontal lineseparating the top and bottom parts.
Proposition 2.3. The pentagram map preserves the Poisson bracket (2.7).
Proof. This is an easy consequence of formula (2.5), see [15] (Lemma 2.9), for the details. 2
Recall that a Poisson structure is a way to associate a vector field to a function. Given afunction f on Pn, the corresponding vector field Xf is called the Hamiltonian vector field definedby Xf (g) = f, g for every function g. In the case of the bracket (2.7), the explicit formula isas follows:
Xf =∑i−j=2
(−1)i+j2 xi xj
(∂f
∂xi
∂
∂xj− ∂f
∂xj
∂
∂xi
). (2.8)
Note that the definitions of the Poisson structure in terms of the bracket of coordinate functions(2.7) and in terms of the Hamiltonian vector fields (2.8) are equivalent.
Geometrically speaking, Hamiltonian vector fields are defined as the image of the map
X : T ∗x Pn → Tx Pn (2.9)
at arbitrary point x ∈ Pn. The kernel of X at a generic point is spanned by the differentials ofthe Casimir functions, that is, the functions that Poisson commute with all functions.
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Remark 2.4. The cluster algebra approach of [7] also provides a Poisson bracket, invariant withrespect to the pentagram map (see the book [5]). It can be checked that this cluster Poissonbracket is induced by the bracket (2.7).
2.5 The rank of the Poisson bracket and the Casimir functions
The corank of a Poisson structure is the dimension of the kernel of the map X in (2.9), thatis, the dimension of the space generated by the differentials of the Casimir functions. In oursituation, in which everything in sight is algebraic, the corank is generically constant.
Proposition 2.5. The Poisson bracket (2.7) has corank 2 if n is odd and corank 4 if n is even;the functions
On = x1x3 · · ·x2n−1, En = x2x4 · · ·x2n (2.10)
for arbitrary n and the functions
On2
=∏
1≤i≤n2
x4i−1 +∏
1≤i≤n2
x4i+1, En2
=∏
1≤i≤n2
x4i +∏
1≤i≤n2
x4i+2, (2.11)
for even n, are the Casimirs of the Poisson bracket (2.7).
Proof. First, one checks that the functions (2.10) and (2.11) are indeed Casimir functions (forarbitrary n and for even n, respectively). To this end, it suffices to consider the brackets of(2.10) and (2.11), if n is even, with the coordinate functions xi.
Second, one checks that the corank of the Poisson bracket is equal to 2, for odd n and 4,for even n. The corank is easily calculated in the coordinates log xi, see [15], Section 2.6 for thedetails. 2
It follows that the Casimir functions are of the form F (On, En), if n is odd, and of the formF (On/2, En/2, On, En), if n is even. In both cases the generic symplectic leaves of the Poissonstructure have dimension 4[(n− 1)/2].
Remark 2.6. If n is even, then the Casimir functions can be written in a more simple manner: ∏1≤i≤n
2
x4i−1,∏
1≤i≤n2
x4i+1,∏
1≤i≤n2
x4i,∏
1≤i≤n2
x4i+2
instead of (2.10) and (2.11).
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2.6 Two constructions of the monodromy invariants
The second main ingredient of the Liouville-Arnold theory is a set of Poisson-commuting invari-ant functions. In this section, we recall the construction [19] of a set of first integrals of thepentagram map
O1, . . . , O[n2 ], On, E1, . . . , E[n2 ], En
called the monodromy invariants. In other words, we will define n+ 1 invariant function on Pn,if n is odd, and n + 2 invariant function on Pn, if n is even. The monodromy invariants arepolynomial in the coordinates (2.4). Algebraic independence of these polynomials was provedin [19]. Note that On and En are the Casimir functions (2.10) and, for even n, the functions On
2
and En2
are as in (2.11).The indexing of the function Oi, Ej corresponds to their weight. More precisely, we define
the weight of the coordinate functions by
|x2i+1| = 1, |x2i| = −1. (2.12)
Then, |Ok| = k and |Ek| = −k. We give two definitions of the monodromy invariants. In [19] itis proved that the two definitions are equivalent.
A. The geometric definition. Given a twisted n-gon (2.1), the corresponding monodromyhas a unique lift to SL3. By slightly abusing notation, we again denote this matrix by M . Thetwo traces, tr(M) and tr(M−1), are preserved by the pentagram map (this is a consequenceof the projective invariance of T ). These traces are rational functions in the corner invariants.Consider the following two functions:
Ω1 = tr(M)O23n E
13n , Ω2 = tr(M−1)O
13n E
23n .
It turns out that Ω1 and Ω2 are polynomials in the corner invariants (see [19]). Since thepentagram map preserves the monodromy, and On and En are invariants, the two functions Ω1
and Ω2 are also invariants. We then have:
Ω1 =
[n/2]∑k=0
Ok, Ω2 =
[n/2]∑k=0
Ek, (2.13)
where Ok has weight k and Ek has weight −k and where we set
O0 = E0 = 1,
for the sake of convenience. The pentagram map preserves each homogeneous component indi-vidually because it commutes with the rescaling (2.6).
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Notice also that, if n is even, then On2
and En2
are precisely the Casimir functions (2.11).However, the invariants On and En do not enter the formula (2.13).
B. The combinatorial definition. Together with the coordinate functions xi, we consider thefollowing “elementary monomials”
Xi := xi−1 xi xi+1, i = 1, . . . , 2n. (2.14)
Let O(X,x) be a monomial of the form
O = Xi1 · · ·Xis xj1 · · ·xjt ,
where i1, . . . , is are even and j1, . . . , jt are odd. Such a monomial is called admissible if thefollowing conditions hold for all relevant indices a, b:
• |ia − ib| > 2.
• |ja − jb| > 4.
• |ia − jb| > 3.
It turns out that this is equivalent to the statement that there are no repeated terms and thatthe Poisson brackets Xir , Xiu and Xir , xju and xjr , xju of all the elementary monomialsentering O vanish.
The weight of the above monomial is
|O| = s+ t,
see (2.12). For every admissible monomial, we also define the sign of O via
sign(O) := (−1)t.
The invariant Ok is defined as the alternated sum of all the admissible monomials of weight k:
Ok =∑|O|=k
sign(O)O, k ∈
1, 2, . . . ,[n
2
]. (2.15)
It is proved in [19] that this definition of Ok coincides with (2.13).
Example 2.7. The first two invariants are:
O1 =n∑i=1
(X2i − x2i+1) , O2 =∑|i−j|≥2
(x2i+1 x2j+1 −X2i x2j+1 +X2iX2j+2) ,
for n ≤ 5 the above formulas simplify, see [15].
The definition of the functions Ek is exactly the same, except that the roles of even and oddare swapped.
Remark 2.8. There is an elegant way to define the monodromy invariants in terms of deter-minants. See [21].
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2.7 The monodromy invariants Poisson commute
The goal is this section is to prove the following result.
Theorem 2. Oi, Oj = Oi, Ej = Ei, Ej = 0 for all relevant indices i and j. Hence, thecorresponding Hamiltonian vector fields commute.
We prove first that Oi, Oj = 0. The only monomials which can appear in Oi, Oj haveexponents in the set 1, 2. In our proof, we sometimes view the monomial µ as a mappingµ : 1, ..., 2n → 0, 1, 2. Here µ(i) is the exponent of xi in µ. The support of µ (as a map) isexactly the set of indices of variables which appear in µ (as a monomial).
For a polynomial P and a monomial µ, we define P |µ as the monomial µ times its coefficientin P . We call µ good if Oi, Oj|µ = 0 for all indices i, j. We will prove that all monomials aregood.
We say that µ decomposes into µ1 and µ2 if (as monomials) µ = µ1µ2, and (as maps) thesupports of µ1 and µ2 are separated by at least 2 empty spaces, in the cyclic sense. It followsthat the Poisson bracket of a variable with the index in support of µ1 and a variable with theindex in support of µ2 vanishes. If we cannot factor µ this way, we call µ indecomposable.
Lemma 2.9. If µ decomposes into µ1 and µ2, and both µ1 and µ2 are good, then µ is good.
Proof. For any monomial F whose support is contained in the support of µ, we have a uniquedecomposition F = F1F2 where F1 (respectively F2) denotes the monomial obtained from F bysetting to 1 all the variables having indices in the support of µ2 (respectively µ1). For example,µ = x1x5x7 decomposes into µ1 = x1 and µ2 = x5x7. If F = x1x5 then F1 = x1 and F2 = x5.
Assume that the support of F is contained in the support of µ. Note that F is an admissiblemonomial in Oi if and only if F1 and F2 are admissible monomials in Oi1 and Oi2 , respectively,where i1 + i2 = i. Furthermore, sign(F ) = sign(F1)sign(F2).
Let Oi be the sum of the terms in Oi whose support is contained in the support of µ, andlikewise for Oj :
where the starred equality is due to the large separation between the supports of A and B andLeibniz’s rule. The last term vanishes because µ1 and µ2 are assumed to be good. We concludethat µ is good as well. 2
Let us record how the Poisson bracket interacts with the elementary monomials. We have
All other brackets Xi, Xj, as well as Xi, xj, vanish.The Poisson bracket Oi, Oj is a sum of monomials of the form
µ = Xi1 · · ·Xisxj1 · · ·xjt , (2.18)
where i1, . . . , is are even and j1, . . . , jt are odd. Using Lemma 2.9, we assume that µ is inde-composable, and we want to prove that µ is good.
If µ contains x2a then either µ = x2a or µ is decomposible. (The follows from the admissibilitycondition.) Likewise, if µ contains X2
a then either µ = X2a or µ is decomposible. These two
“singleton” cases are trivial, so henceforth we will assume that µ does not contain the square ofan elementary monomial.
Define an oriented graph Γµ whose vertex set is Xi1 , . . . , Xis , xj1 , . . . , xjt, the elementarymonomials which appear in µ. Join vertex v1 with v2 by an oriented edge from v1 to v2 ifv1, v2 = v1v2. Recall that all the non-zero brackets of elementary monomials are listed in(2.16) and (2.17), and they all have coefficients ±1.
Note that if Γµ is disconnected then µ is decomposable. Hence Γµ is a connected graph.
Lemma 2.10. Γµ has no 3-cliques (i.e., triangles), no cycles of odd length, and no verticeshaving in-degree, or out-degree, greater than 1.
Proof. (i) Assume there is a triangle (a, b, c) of elementary monomials. Then two of the threeinvolved elementary monomials belong to the decomposition of the same polynomial, eitherOi or Oj ; assume that a, b ∈ Oi. The monomials a and b are joined by an arrow, thus theirPoisson bracket does not vanish. This leads to a contradiction since all the monomials in Oi areadmissible (see Section 2.6, definition B).
(ii) The argument for odd cycles is the same as in (i): the monomials in the cycle shouldalternate between Oi or Oj , and since the cycle is odd, two adjacent monomials will land on thesame Oi.
(iii) We will show that no vertex of Γµ has out-degree greater than 1. The in-degree caseis similar. Suppose that some vertex Xi has at least 2 outgoing arrows. Since the indices i
15
are even and the indices j are odd, Xi can be joined by an outgoing arrow to the followingvertices (provided they belong to the graph): Xi−2, Xi−4, xi−1, xi−3. Indeed, the coefficient ofthe bracket in all these cases is +1. However, any two vertices from this list, would also bejoined together by an edge. This would give us a triangle, which we have already ruled out. Asimilar (but simpler) consideration applies to a vertex xi: it can be joined by outgoing arrowsto Xi−1 and Xi−1 which are connected by an edge. The argument for two incoming arrows issimilar. 2
It follows that Γµ is of the form:
a1 −−−−→ a2 −−−−→ a3 −−−−→ · · · −−−−→ a`, (2.19)
or Γµ is a cycle of even length; here ai are elementary monomials appearing in (2.18).Suppose ` is odd and assume that i > j. Let A be an admissible monomial in Oi and B
that in Oj such that AB is a scalar multiple of µ (otherwise, trivially, A,B|µ = 0). By theadmissibility condition, A = a1a3 . . . a` and B = a2a4 . . . a`−1. By the Leibniz rule,
A,B =
(a1, a2a1a2
+a3, a2a2a3
+a3, a4a3a4
+ · · ·)AB = (1− 1 + 1∓ · · · )AB = 0 (2.20)
since the number of terms is even. Alternatively, one notices that the intersection number of Γµwith the separating line of the bibartite graph is zero; see Section 2.4.
Suppose now that ` is even. Again let A be an admissible monomial in Oi and B that inOj such that AB is a scalar multiple of µ. Then either A = a1a3 . . . a`−1 and B = a2a4 . . . a`,or the other way around. In both cases, i = j = `/2, contradicting the assumption that i > j.The same argument works if Γµ is a cycle of even length: the vertices must alternate betweenA and B, hence the weights of A and B are equal and thus i = j, a contradiction.
The same proof (or symmetry) shows that Ei, Ej = 0 as well.
To prove that Oi, Ej = 0, we will start out with the same set-up as above. Lemma 2.9works again, and gets us to the indecomposible case. This time we will analyze the structure ofan indecomposible µ without the aid of the bipartite graph. As above, we sometimes think of µas a map from 1, ..., 2n to 0, 1, 2.
Lemma 2.11. No terms contribute to Oi, Ej|µ unless µ has the form
xαi (Xi+3Xi+6Xi+9 · · ·Xi+3`)xβi+3`+2, (2.21)
where α, β ∈ 0, 1. (This way of expressing µ is not necessarily unique.)
16
Proof. Let (A,B) be a supposedly contributing pair. Suppose µ contains the variables xa−1and xa+1 but not xa. If a is even (respectively odd), then B (respectively A), has an odd-indexed(respectively even-indexed) variable but not both the adjacent even-indexed (respectively odd-indexed) variables. This is a contradiction. This shows that the support of µ is a consecutivestring of indices.
Now suppose that µ has the square of some variable. If the support of µ is contained in3 indices, the result is trivial, so we suppose otherwise. Suppose that µ contains the squareof an odd-indexed variable. Say µ contains x25. Then B contains x4x5x6 but not xk whenk ∈ 1, 2, 3, 7, 8, 9. Moreover, A contains x5, but not both x4 and x6. If neither x4 nor x6appears in A, then xk does not appear in A for k ∈ 2, 3, 4, 6, 7, 8. But then µ does not containxk for k = 2, 3, 7, 8. Hence µ is decomposible. If x4 appears in A, then x3x4x5 appears in A andµ does not contain xk for k ∈ 0, 1, 2, 6, 7, 8. But then xk does not appear in µ for k = 1, 2, 7, 8and µ is decomposible. The proof is the same when A contains x6. A similar argument workswhen, µ contains the square of an even-indexed variable.
Now we know that µ is a consecutive string of indices and µ takes on the value 1 on any indexin its support. The admissibility of A and B now forces the structure claimed in the lemma. 2
Just as in the odd-odd case, the way the elementary monomials are assigned to A and Balternates – this follows from parity considerations. Using Leibniz’s rule, we see that A,B = 0unless α + β = 1. In this case, there are exactly two ways to express µ as in Equation 2.21and correspondingly there are exactly two terms contributing to Oi, Oj|µ and they cancel. Toillustrate this principle, we will consider the example where i = 2 and j = 1 and ` = 2. Aftersuitably shifting the indices, we have
µ = x1x2x3x4x5x6x7 = x1X3X6 = X2X5x7,
the two terms x1X6, X3 and X2x7, X5 cancel.This completes the proof that Oi, Ej = 0.
In [19] it is proved that the monodromy invariants are algebraically independent. The argu-ment is rather complicated, but it is very similar in spirit to the related independence proof wegive in Section 4. The algebraic independence result combines with Theorem 2 to establish theintegrability of the pentagram map on the space Pn. Indeed, the Poisson bracket (2.7) defines asymplectic foliation on Pn, the symplectic leaves being locally described as levels of the Casimirfunctions, see Proposition 2.5. The number of the remaining invariants is exactly half of thedimension of the symplectic leaves. The classical Liouville-Arnold theorem [1] is then applied.
3 Integrability on Cn modulo a calculation
The general plan of the proof of Theorem 1 is as follows.
17
1. We show that the Hamiltonian vector fields on Pn corresponding to the monodromy in-variants are tangent to the subspace Cn,
2. We restrict the monodromy invariants to Cn and show that the dimension of a generic levelset is n− 4 if n is odd and n− 5 if n is even.
3. We show that there are exactly the same number of independent Hamiltonian vector fields.
In this section, we prove the first statement and also show that the dimension of the level setsis at most n− 4 if n is odd and n− 5 if n is even, and similarly for the number of independentHamiltonian vector fields. The final step of the proof that this upper bound is actually thelower one will be done in the next two sections. This final step is a nontrivial calculation thatcomprises the bulk of the paper.
3.1 The Hamiltonian vector fields are tangent to CnThe space Cn is a subvariety of Pn having codimension 8. It turns out that one can give explicitequations for this variety. See Lemma 5.3. (These equations do not play a role in our proof, butthey are useful to have.)
The following statement is essentially a consequence of Theorem 2. This is an importantstep of the proof of Theorem 1.
Proposition 3.1. The Hamiltonian vector field on Pn corresponding to a monodromy invariantis tangent to Cn.
Proof. The space Pn is foliated by isomonodromic submanifolds that are generically of codi-mension 2 and are defined by the condition that the monodromy has fixed eigenvalues. Hencethe isomonodromic submanifolds can be defined as the level surfaces of two functions, tr(M)and tr(M−1). This foliation is singular, and Cn is a singular leaf of codimension 8. We notethat the versal deformation of Cn is locally isomorphic to SL(3) partitioned into the conjugacyequivalence classes.
Consider a monodromy invariant, F (= Oi or Ei), and its Hamiltonian vector field, XF .We know that the Poisson bracket F, tr(M) = 0, since all monodromy invariants Poissoncommute and tr(M) is a sum of monodromy invariants. Hence XF is tangent to the genericleaves of the isomonodromic foliation on Pn. Let us show that XF is tangent to Cn as well.
In a nutshell, this follows from the observation that the tangent space to Cn at a smoothpoint x0 is the intersection of the limiting positions of the tangent spaces to the isomonodromicleaves at points x as x tends to x0. Assume then that XF is transverse to Cn at point x0 ∈ Cn.Then XF will be also transverse to an isomonodromic leaf at some point x close to x0, yieldinga contradiction.
More precisely, we can apply a projective transformation so that the vertices V1, V2, V3, V4of a twisted n-gon V1, V2, . . . become the vertices of a standard square. This gives a local
18
identification of Pn with the set of tuples (V5, . . . , Vn;M) where M is the monodromy, theprojective transformation that takes the quadruple (V1, V2, V3, V4) to (Vn+1, Vn+2, Vn+3, Vn+4).The space of closed n-gons is characterized by the condition that M is the identity. Thus wehave locally identified Pn with Cn × SL(3). In particular, we have a projection Pn → SL(3),and the preimage of the identity is Cn. The isomonodromic leaves project to the conjugacyequivalent classes in SL(3).
Thus our proof reduces to the following fact about the group SL(3) (which holds for SL(n)as well).
Lemma 3.2. Consider the singular foliation of SL(3) by the conjugacy equivalence classes, andlet TX be the tangent space to this foliation at X ∈ SL(3). Then the intersection, over all X, ofthe limiting positions of the spaces TX , as X → 11, is trivial (here 11 ∈ SL(3) is the identity).
Proof. Let B ∈ SL(3), and let B + εC be an infinitesimal deformation within the conjugacyequivalence class. Then
tr (B + εC) = tr(B), tr((B + εC)2
)= tr
(B2),
hence tr(C) = 0 and tr(BC) = 0, and also tr(B−1C) = 0 since det(B + εC) = 1. Thus thetangent space to a conjugacy equivalent class of B is given by
tr(C) = tr(BC) = tr(B−1C) = 0.
Now let B = 11 + εA, a point in an infinitesimal neighborhood of the identity 11; we havetr(A) = 0. Then our conditions on C implies tr(C) = tr(AC) = 0. Since tr(AC) is a non-degenerate quadratic form, an element C ∈ sl(3) satisfying tr(AC) = 0 for all A ∈ sl(3) has tobe zero. 2
In view of what we said above, this implies the proposition. 2
3.2 Identities between the monodromy invariants
In this section, we consider the restriction of the monodromy invariants from the space of alltwisted n-gons to the space Cn of closed n-gons. We show that these restrictions satisfy 5non-trivial relations, whereas their differentials, considered as covectors in Pn whose foot-pointsbelong to Cn, satisfy 3 non-trivial relations. These relations are also mentioned in [15] and [22].In Sections 4 and 5, we will prove that there are no other relations between the monodromyinvariants on Cn and their differentials along Cn.
We remark that, strictly speaking, the identities established in this section are not neededfor the proof of our main result. For the main result, all we need to know is that there areenough commuting flows to fill out what could be (a priori , without the results in this section)
19
a union of level sets of the monodromy invariants. Thus, the reader interested only in the mainresult can skip this section.
Theorem 3. (i) The restrictions of the monodromy integrals to Cn satisfy the following fiveidentities:
[n/2]∑j=0
Oj = 3E13n O
23n ,
[n/2]∑j=0
Ej = 3E23n O
13n ,
[n/2]∑j=1
j Oj = nE13n O
23n ,
[n/2]∑j=1
j Ej = nE23n O
13n ,
E13n
[n/2]∑j=1
j2Oj = O13n
[n/2]∑j=1
j2Ej .
(3.1)
(ii) The differentials of the monodromy integrals along Cn satisfy the three identities:
[n/2]∑j=1
dOj = 2E13n O
− 13
n dOn + E− 2
3n O
23n dEn,
[n/2]∑j=1
dEj = 2E− 1
3n O
13n dEn + E
23n O
− 23
n dOn,
O13n
( [n/2]∑j=1
j dEj
)+ E
13n
( [n/2]∑j=1
j dOj
)= nE
23nO
23n
(E−1n dEn +O−1n dOn
).
(3.2)
Proof. Recall that the monodromy invariants Oj are the homogeneous components of the
polynomial O2/3n E
1/3n tr(M) with respect to the rescaling (2.6), where s = et for convenience.
Likewise, the monodromy invariants Ej are homogeneous components of O1/3n E
2/3n tr(M−1).
Recall also that O0 = E0 = 1.Denote for simplicity O
1/3n E
2/3n = U, O
2/3n E
1/3n = V . Notice that the monodromy matrix M
has the unit determinant. Let eλ1 , eλ2 , eλ2 be the eigenvalues of M . One has
λ1 + λ2 + λ3 ≡ 0. (3.3)
We consider a one-parameter family of n-gons depending on the rescaling parameter t, such thatfor t = 0, the n-gon belongs to Cn. The monodromy M = Mt also depends on t so that wethink of λi as functions of the corner coordinates (x1, . . . , x2n) and of t. For t = 0, one has:λi = 0, i = 1, 2, 3 since M0 = Id.
20
The eigenvalues of M−1 are e−λ1 , e−λ2 , e−λ2 . Since the weights of Oi and Ej are j and −jrespectively, the definition of the integrals writes as follows:
ent3 V
(eλ1 + eλ2 + eλ2
)=
[n/2]∑j=0
etj Oj , e−nt3 U
(e−λ1 + e−λ2 + e−λ2
)=
[n/2]∑j=0
e−tj Ej .
which we rewrite as
V(eλ1 + eλ2 + eλ2
)=
[n/2]∑j=0
et(j−n3)Oj , U
(e−λ1 + e−λ2 + e−λ2
)=
[n/2]∑j=0
e−t(j−n3)Ej . (3.4)
Setting t = 0 in these formulas yields the first two identities in (3.1). Next, differentiatethese equations in t :
V3∑i=1
λ′i eλi =
[n/2]∑j=0
(j − n
3
)etj Oj ,
where λ′i = dλi/dt, and similarly for Ej . Set t = 0, then the left-hand-side vanishes because∑λ′i = 0 due to (3.3). Hence
[n/2]∑j=0
j Oj =n
3
[n/2]∑j=0
Oj = nV
due to the first identity in (3.1) and similarly for Ej . One thus obtains the third and the fourthidentity in (3.1).
To obtain the fifth equation in (3.1), differentiate the equations (3.4) with respect to t twiceto get
V( 3∑i=1
(λ′′i + λ′2i ) eλi)
=
[n/2]∑j=0
(j − n
3
)2etj Oj ,
U( 3∑i=1
(−λ′′i + λ′2i
)eλi)
=
[n/2]∑j=0
(j − n
3
)2e−tj Ej .
Divide the first equality by V , the second by U , subtract one from another, and set t = 0:
23∑i=1
λ′′i = V −1[n/2]∑j=0
(j − n
3
)2Oj − U−1
[n/2]∑j=0
(j − n
3
)2Ej .
The left hand side vanishes, due to (3.3), so
V −1[n/2]∑j=0
(j − n
3
)2Oj = U−1
[n/2]∑j=0
(j − n
3
)2Ej . (3.5)
21
Therefore
V −1[n/2]∑j=0
j2Oj −2n
3V −1
[n/2]∑j=0
j Oj + V −1n2
9
[n/2]∑j=0
Oj =
U−1[n/2]∑j=0
j2Ej −2n
3U−1
[n/2]∑j=0
j Ej + U−1n2
9
[n/2]∑j=0
Ej .
The second and the third terms on the left and the right hand sides are pairwise equal, due tothe first four identities in (3.1). This implies the fifth identity (3.1).
To prove (3.2), take differentials of (3.4):
V
3∑i=1
eλi dλi +( 3∑i=1
eλi)dV =
( [n/2]∑j=0
(j − n
3
)et(j−
n3)Oj
)dt+
[n/2]∑j=0
et(j−n3) dOj ,
and
−U3∑i=1
e−λi dλi +( 3∑i=1
e−λi)dU =
−( [n/2]∑j=0
(j − n
3
)e−t(j−
n3)Ej
)dt+
[n/2]∑j=0
e−t(j−n3) dEj .
Set t = 0: the first terms on the right hand sides vanish due to (3.3), and the first parentheseson the right hand sides vanish due to (3.1). We get
[n/2]∑j=0
dOj = 3 dV,
[n/2]∑j=0
dEj = 3 dU,
the first two identities in (3.2).Finally, differentiate the above equations with respect to t and set t = 0 to obtain:
V3∑i=1
λ′i dλi + V3∑i=1
d(λ′i) +( 3∑i=1
λ′i
)dV =
( [n/2]∑j=0
(j − n
3
)2Oj
)dt+
[n/2]∑j=0
(j − n
3
)dOj ,
U3∑i=1
λ′i dλi − U3∑i=1
d(λ′i) +( 3∑i=1
λ′i
)dU =
( [n/2]∑j=0
(j − n
3
)2Ej
)dt−
[n/2]∑j=0
(j − n
3
)dEj .
22
Once again, the second and the third sums on the left hand sides vanish, due to (3.3). Dividethe first equation by V , the second by U , and subtract one from another, using (3.5):
V −1[n/2]∑j=0
(j − n
3
)dOj + U−1
[n/2]∑j=0
(j − n
3
)dEj = 0.
Hence
V −1[n/2]∑j=0
j dOj + U−1[n/2]∑j=0
j dEj =n
3
(V −1
[n/2]∑j=0
dOj + U−1[n/2]∑j=0
dEj
).
Due to the first two identities in (3.2), the right-hand-side equals n (O−1n dOn +E−1n dEn). Thisyields the third identity in (3.2). Theorem 3 is proved. 2
Remark 3.3. a) Let E be the Euler vector field that generates the scaling. Then
E(Oj) = j Oj , E(Ej) = −j Ej .
If one evaluates the differentials in the identities (3.2) on E , one obtains the last three identitiesin (3.1). This is a check that (3.1) and (3.2) are consistent with each other.
b) Equivalently, (3.2) can be rewritten as
3 dOn = 2E− 1
3n O
13n
(∑[n/2]j=1 dOj
)− E−
23
n O23n
(∑[n/2]j=1 dEj
),
3 dEn = 2E13n O
− 13
n
(∑[n/2]j=1 dEj
)− E
23n O
− 23
n
(∑[n/2]j=1 dOj
),
0 = O13n
(3
[n/2]∑j=1
j dEj − n[n/2]∑j=1
dEj
)+ E
13n
(3
[n/2]∑j=1
j dOj − n[n/2]∑j=1
dOj
).
c) The identities (3.1) and (3.2) are satisfied in a larger subspace than Cn, consisting oftwisted polygons whose monodromy has equal eigenvalues. This subspace has codimension 2in Pn.
d) In both cases, n odd and n even, the kernel of the Poisson map X (2.9) (spanned by thedifferentials of the Casimir functions) has zero intersection with the subspace of T ∗Pn spannedby the relations 3.2.
3.3 Reducing the proof to a one-point computation
For ease of exposition, we will give our proof only in the odd case, and we set n ≥ 7 odd. Modulochanging some of the indices, the even case is similar. We will explain everything in terms ofthe odd case and, at the end of this section, briefly explain what happens in the even case.
23
Let M denote the algebra generated by the monodromy invariants. In the Section 4 wemake the following calculations.
1. There exist elements F1, ..., Fn−2 ∈ M and a point p ∈ Cn such that the differentialsdF1, ..., dFn−2 are linearly independent at p. Therefore, dF1, ..., dFn−2 are linearly inde-pendent at almost all q ∈ Cn.
2. There exist elements G1, ..., Gn−4 ∈ M and a point p ∈ Cn such that the differentialsdG1|TpCn , ..., dGn−4|TpCn are linearly independent. Therefore, dG1|TqCn , ..., dGn−4|TqCn arelinearly independent at almost all q ∈ Cn.
In Calculation 1, we are computing the differentials on the ambient space Pn but evaluatingthem at a point of Cn. In Calculation 2, we are computing the differentials on the ambient space,evaluating them at a point of Cn, and restricting the resulting linear functionals to the tangentspace of Cn. In both calculations, we are actually evaluating at points in C0
n. In each case,what allows us to make a conclusion about generic points is that the monodromy invariants arealgebraic.
Calculation 2 combines with Theorem 3 to show that there are exactly n − 4 algebraicallyindependent monodromy invariants, when restricted to Cn. Hence, the generic common level setof the monodromy invariants Oi, Ei, restricted to Cn, has dimension n− 4.
Next, we wish to prove that these level sets have locally free action of the abelian group Rd(or Cd in the complex case). For F ∈ M, the Hamiltonian vector field XF is tangent to Cn,by Proposition 3.1, and also tangent to the common level set of functions in M. Finally, byTheorem 2, the Hamiltonian vector fields all commute with each other (i.e., define an action ofthe Abelian Lie algebra). The following lemma finishes our proof.
Lemma 3.4. The Hamiltonian vector fields of the monodromy invariants generically span themonodromy level sets on Cn.
Proof. Let ∧1Pn denote the space of 1-forms on Pn. Let X denote the space of vector fieldson Cn. Let dM⊂ ∧1Pn denote the image ofM under the d-operator. Calculation 1 shows thatthe vector space dM generically has dimension n − 2 when evaluated at points of Cn. At thesame time, we have the Poisson map X : dM→ X , given by
X(dF ) = XF ,
see (2.9). In the odd case, the map X has 2 dimensional kernel, see Remark 3.3 d). Hence, Xhas n− 4 dimensional image, as desired. 2
Now we explain explicitly how the results above give us the quasi-periodic motion in thecase of closed convex polygons. We know from the work in [17] that the monodromy level sets
24
on C0n are compact. By Sard’s Theorem, and by the calculations above, almost every level setis a smooth compact manifold of dimension m = n − 4. By Sard’s Theorem again, and bythe dimension count above, almost every level set L possesses a framing by Hamiltonian vectorfields. That is, there are m Hamiltonian vector fields on L which are linearly independent ateach point and which define commuting flows. These vector fields define local coordinate chartsfrom L into Rm, such that the overlap functions are translations. Therefore L is a finite unionof affine m-dimensional tori. The whole structure is invariant under the pentagram map, andso the pentagram map is a translation of L relative to the affine structure on L. This is thequasi-periodic motion. Even more explicitly, some finite power of the pentagram map preserveseach connected component of L and is a constant shift on each connected component.
The Even Case: In the even case, we have the following calculations:
1. There exist elements F1, ..., Fn−1 ∈ M and a point p ∈ Cn such that the differentialsdF, ..., dFn−1 are linearly independent at p. Therefore, dF, ..., dFn−1 are linearly indepen-dent at almost all q ∈ Cn.
2. There exists elements G1, ..., Gn−3 ∈ M and a point p ∈ Cn such that the differentialsdG1|TpCn , ..., dGn−3|TpCn are linearly independent. Therefore, dG1|TpCn , ..., dGn−3|TpCn arelinearly independent at almost all q ∈ Cn.
In this case, the common level sets generically have dimension n−5 and, again, the Hamilto-nian vector fields generically span these level sets. The situation is summarized in the followingtable.
Invariants Casimirs Level sets /Hamiltonian fieldsn odd n+ 1 2 d = n− 4n even n+ 2 4 d = n− 5
4 The linear independence calculation
4.1 Overview
For any given (smallish) value of n, one can make the calculations directly, at a random point,and see that it works. The difficulty is that we need to make one calculation for each n. Onemight say that the idea behind our calculations is tropicalization. The monodromy invariantsand their gradients are polynomials with an enormous number of terms. We only need to makeour calculation at one point, but we will consider a 1-parameter family of points, depending ona parameter u. As u→ 0, the different variables tend to 0 at different rates. This sets up a kindof hierarchy (or filtration) on the the monomials comprising the polynomials of interest to us,and only the “heftiest” monomials in this hierarchy matter. This reduces the whole problem toa combinatorial exercise.
25
We take n ≥ 7 odd. Let m = (n− 1)/2. Recall that M is spanned by
O1, ..., Om, On, E1, ..., Em, En.
We defineAk,± = Ok ± Ek. (4.1)
For the first calculation, we use the monodromy invariants
For the second calculation, we use the monodromy invariants
A3,−, ..., Am,−, A3,+, ..., Am,+, An,+. (4.3)
The point we use is of the form p = P u, where P u is an n-gon having corner invariants
a, b, c, d, u1, u2, u3, u4, ..., u4, u3, u2, u1, d, c, b, a, (4.4)
Here
• a = O(u(n−4)(n−3)/2).
• b = 1 +O(u)
• c = 1 +O(u).
• d = 1 +O(u).
We will show that the results hold when u is sufficiently small. Here we are using the big Onotation, so that O(u) represents an expression that is at most Cu in size, for a constant C thatdoes not depend on u.
We will construct P u in the next section. Our first calculation requires only the informationpresented above. The second calculation, which is almost exactly the same as the first calcula-tion, requires some auxilliary justification. In order to justify the calculation we make, we needto make some estimates on the tangent space TPu to Cn at P u. We will also do this in the nextsection.
In Section 4.2 and Section 4.3 we will explain our two calculations in general terms. InSection 4.4 we will define the concept of the heft of a monomial, and we will use this concept toput a kind of ordering on the monomials that appear in the monodromy invariants of interestto us. Following the analysis of the heft, we complete the details of our calculations.
26
4.2 The first calculation in broad terms
Let ∇ denote the gradient on R2n. Let ∇ denote the normalized gradient :
∇F = λ−1∇F ; λ = ‖∇F‖∞. (4.5)
In practice, we never end up dividing by zero. So, the largest entry in ∇F is ±1.If F is a monodromy invariant, the coordinates of ∇F (P u) have a power series in u. We
define ΨF to be the result of setting all terms except the constant term to 0. We call ΨF theasymptotic gradient . Thus, if
Lemma 4.1. Suppose that ΨF1, ...,ΨFk are linearly independent. Then likewise ∇F1, ...,∇Fkare linearly independent at P u for u sufficiently small. Equivalently, the same goes for dF1, ..., dFk.
Proof. Since ΨF1, ...,ΨFk are independent there is some ε > 0 such that a sum of the form∣∣∣∣∑ bjΨFj
∣∣∣∣ < ε; max |bj | = 1
is impossible.Suppose for the sake of contradiction that the gradients are linearly dependent at P u for
all sufficiently small u. Then the normalized gradients are also linearly dependent at P u for allsufficiently small u. We may write∑
bj∇Fj · ei = 0; max |bj | = 1. (4.6)
for the standard basis vectors e1, ..., e2n. The coefficients bj possibly depend on u, but thisdoesn’t bother us.
We have the bound ∣∣∣∣bj∇Fj − bjΨFj∣∣∣∣ = O(u). (4.7)
Hence ∑j
bjΨFj · ei = O(u) (4.8)
for all basis vectors ei. Therefore, we can take u small enough so that∣∣∣∣∑ bjΨFj
∣∣∣∣ < ε; max |bj | = 1,
in contradiction to what we said at the beginning of the proof. 2
27
Remark 4.2. The idea of the proof of the previous lemma is simple: given a matrix, alge-braically dependent on a parameter u, the rank of the matrix is greatest in a Zariski open subsetof the parameter space and can only drop for special values of the parameter (zero, in our case).
We form a matrix M+ whose rows are ΨF , where F is each of the A+ invariants. We similarlyform the matrix M−.
Lemma 4.3. Each row of M+ is orthogonal to each row of M−.
Proof. Consider the map T : R2n → R2n which simply reverses the coordinates. We haveEk T = Ok for all k and moreover T (P u) = P u. Letting dT be the differential of T , we have
dT (∇Ak,±) = ±∇Ak,±. (4.9)
Our lemma follows immediately from this equation, and from the fact that T is an isometricinvolution. 2
In view of Lemmas 4.1 and Lemma 4.3, our first calculation follows from the statements thatM+ and M− have full rank.
For the matrix M+, we consider the minor m+ consisting of columns
1, 6, 7, 10, 11, 14, 15, 18, 19, ...
until we have a square matrix. We will prove below that m+ has the following form (shown inthe case n = 13.)
This matrix always has full rank. Hence M+ has full rank.For the matrix M− we consider the minor m− consisting of columns
1,3, 6, 7, 10, 11, 14, 15, 18, 19, ...
The only difference here is that column 3 is inserted. The resulting matrix has exactly the samestructure as just described. Hence M− has full rank.
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4.3 The second calculation in broad terms
Let T = TPu(Cn) denote the tangent space to Cn at P u. Let ek denote the standard basis forR2n. Let π : R2n → R2n−8 denote the map which strips off the first and last 4 coordinates.
Define∇8 = π ∇. (4.11)
We define the normalized version ∇8 exactly as we defined ∇. Likewise we define Ψ8G for anymonodromy function G.
For a collection of vectors v5, . . . , v2n−4 to be specified in the next lemma, we form the vector
Υ8G = (Dv5G, ...,Dv2n−4G) (4.12)
made from the directional derivatives of G along these vectors. Note, by way of analogy, that
∇8G = (De5G, ...,De2n−4G). (4.13)
We define the normalized version Υ8 exactly as we defined ∇8.In the next section, we will establish the following result.
Lemma 4.4 (Justification). There is a basic v5, ..., v2n−4 for TPu(Cn) such that π(vk) = ek forall k and
Υ8G− ∇8G = O(u).
Corollary 4.5. Suppose that Ψ8G1, ...,Ψ8Gk are linearly independent. Then the restrictions ofdG1, ..., dGk to TPu(Cn) are linearly independent for u sufficiently small.
Proof. Given our basis, Ψ8 represents the constant term approximation of both Υ8 and ∇8.So, the same proof as in Lemma 4.1 shows that the vectors Υ8Gj are linearly independent. Thisis equivalent to the conclusion of our corollary. 2
Using the invariants listed in (4.3), we form the matrices M+ and M− just as above, usingΨ8 in place of Ψ. Lemma 4.3 again shows that each row of M+ is orthogonal to each row of M−.Hence, we can finish the second calculation by showing that both M+ and M− have full rank.
For M− we create a square minor m− using the columns
2, 3, 6, 7, 10, 11, 14, 15, ...
Again, we continue until we have a square. It turns out that m− has the form±1 ±1 ±1 ±10 ±1 ±1 ±10 0 ±1 ±10 0 0 ±1
(4.14)
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Hence M− has full rank.For M+ we create a square minor m+ using the same columns, but extending out one further
(on account of the larger matrix size.) It turns out that m+ has the same form as m−. HenceM+ has full rank.
4.4 The heft
Any monomial in the variables x1, ..., x2n, when evaluated at P u, has a power series expansionin u. We define the heft of the monomial to be the smallest exponent that appears in this series.For instance, the heft of u2 + u3 is 2. We define the heft of a polynomial to be the minimumheft of the monomials that comprise it. Given a polynomial F , we define heft of ∇F to be theminimum heft, taken over all partial derivatives ∂F/∂xj .
We call a monomial term of ∂F/∂xk hefty if its heft realizes the heft of ∇F . We define HkFto be the sum of the hefty monomials in ∂F/∂xk. Each monomial occurs with sign ±1. Wedefine |HkF | ∈ Z to be the sum of the coefficients of the hefty terms in HkF . We say that F isgood if |HkF | 6= 0 for at least one index k. If F is good then
ΨF = C(|H1F |, ..., |H2nF |), (4.15)
for some nonzero constant C that depends on F . It turns out that C = ±1 in all cases.We say that F is great if F is good and |HkF | 6= 0 for at least one index k which is not
amongst the first or last 4 indices. When F is great, not only does equation (4.15) hold, but wealso have
Ψ8F = C(|H5F |, ..., |H2n−4F |), (4.16)
Lemma 4.6. Let k = 2, 3. Then Ak,± is great and ∇Ak,± has heft 0.
Proof. Let F = Ak,±. Consider the case k = 2. The argument turns out to be the same inthe (+) and (−) cases. We say that an outer variable is one of the first or last 4 variables inR2n, and we call the remaining variables inner . Since x2x6 and x6x2n−2 are both terms of F ,we see that
H6F = x2 + x2n−2 + ...
In particular, ∇F has heft 0. Any term in H6F involves only the outer 8 variables, and a shortcase-by-case analysis shows that there are no other possibilities besides the two terms listedabove. Hence |H6F | = 2. This shows that F is great.
Now consider the case k = 3. The argument turns out to be the same in the (+) and (−)cases. Since x2x6x2n−2 is a term of F we see that
H6F = x2x2n−2 + ...
The rest of the proof is as in the previous case, with the only difference being that |H6F | = 1in this case. 2
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From now on, we fix some F = Ak,± with 3 < k ≤ m. Let α1, α2, ... be the terms of thefollowing sequence
0, 0, 0, 2, 3, 6, 7, 10, 11, 14, 15, ... (4.17)
Lemma 4.7. ∇F has heft at most α1 + ...+ αk.
Proof. We describe a specific term in ∇F having heft α1 + ... + αk. We make a monomialusing the indices
2, 2n− 2, 6, 2n− 6, 10, 2n− 10, ... (4.18)
stopping when we have used k − 1 numbers. The monomial corresponding to these indices hasheft
Thinking of our indices cyclically, we see that our integers lie in an interval of length 4k − 7.So, between the largest index in (4.18) that is less than n and the smallest index greater thann there is an unoccupied stretch of at least 9 integers. The point here is that
9 + (4k − 7) ≤ 9 + 4m− 7 = 9 + 2(n− 1)− 7 = 2n.
Given that the unoccupied stretch has at least 9 consecutive integers, there is at least 1 (andin fact at least 2) even indices j such that the monomial
m = ±xjx2x2n−2x6x2n−6x10...
is a term of F . But then ∂m/∂xj has heft α1 + ...+ αk. 2
We mention that (4.18) is one of two obvious ways to make a term of heft α1 + ...+αk. Theother way is to take the mirror image, namely
2n− 1, 3, 2n− 5, 7, 2n− 9, 11, ... (4.19)
Lemma 4.8. If ∂F/∂xj has a hefty term, then j is an inner variable.
Proof. For ease of exposition, we will consider the case when j is one of the first 4 variables. Let(i1, ..., id) be the sequence of indices which appear in a term m′ of ∂F/∂xj . The correspondingterm m in F has index sequence (j, i1, ..., id), where these numbers are not necessarily writtenin order. We know that at least one of the indices, say a, is an inner variable. By construction∂m/∂xa has smaller heft than m′. Hence ∂F/∂xj has no hefty terms. Hence j is an innervariable. 2
Lemma 4.9. Suppose the monomial ±xi1 ...xia is a hefty term of ∂F/∂xj. Then a = k− 1 andi1, ..., ik−1 are either as in (4.18) or as in equation (4.19).
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Proof. We have to play the following game: We have a grid of 2n dots. The first and last dotare labelled (n − 3)(n − 4)/2. The remaining 6 outer dots are labelled 0. The inner dots arelabelled 1, 2, 3, ..., 3, 2, 1. Say that a block is a collection of d dots in a row for d = 1, 2, 3. Wemust pick out either k or k − 1 blocks in such a way that the total sum of the correspondingdots is as small as possible, and the (cyclically reckoned) spacing between consecutive blocks isat least 4. That is, at least 3 “unoccupied dots” must appear between every two blocks.
It is easy to see that one should use k − 1 blocks, all having size 1. Moreover, half (or halfminus one) of the blocks should crowd as much as possible to the left and half minus one (orhalf) of the blocks should crowd as much as possible to the right. A short case by case analysisof the placement of the first and last blocks shows that one must have precisely the choices madein (4.18) and (4.19). 2
Corollary 4.10. Let F = Ak,±, with k ≥ 2. Then F is good. If k ≤ m then F is great, and theheft of ∇F is α1 + ...+ αk.
Proof. In light of the results above, the only nontrivial result is that F is great when 3 < k ≤m. The construction in connection with (4.18) produces a hefty term of ∂F/∂xj for some innerindex j. The key observation is that, for parity considerations, the mirror term correspondingto (4.19) is not a term of ∂F/∂xj . In one case j must be odd and in the other case j must beeven. Hence, there is only 1 hefty term in ∂F/∂xj . 2
As regards the heft, we have done everything but analyze the Casimirs. Recall that
On = x1x3...x2n−1; En = x2x4...x2n. (4.20)
Lemma 4.11. An,± is good and ∇An,+ has heft
(n− 3)(n− 4)
2.
Moreover,ΨAn,± = (1, 0, ..., 0,±1).
Proof. Let F be either of these functions. Clearly the hefty terms of ∇F are the ones whichomit the first and last variables. From here, this lemma is an exercise in arithmetic. 2
A similar argument proves
Lemma 4.12. An,± is good and ∇8An,+ has heft (n− 4)2. Moreover,
ΨAn,± = (0, ..., 0,±1, 1, 0, ..., 0),
with the 2 middle indices being nonzero.
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4.5 Completion of the first calculation
To complete the first calculation, we need to analyze the matrix made from the asymptoticgradients ΨF1,ΨF2, .... We deal with the first two in a calculational way.
Proof. Let F = A2,±. We know that F has heft 0, so the hefty terms in ∇F are monomialswhich only involve the outer indices. Hence, when 8 ≤ j ≤ 2n − 8 the result only depends onthe parity of j and neither the value of j nor the value of n. For the remaining indices, theresult is also independent of n. Thus, a calculation in the case (say) n = 13 is general enoughto rigorously establish the whole pattern. This is what we did. 2
Now we are ready to analyze the minors m+ and m− described in connection with the firstcalculation. When we say that a certain part of one of these matrices has the form given by(4.10), we understand that (4.10) gives a smallish member of an infinite family of matrices, allhaving the same general type. So, we mean to take the corresponding member of this familywhich has the correct size.
We say that a given row or column of one of our matrices checks if it matches the form givenby (4.10). We will give the argument for m+. The case for m− is essentially the same.
Lemma 4.15. The first column of m+ checks.
Proof. By Lemma 4.11, the first coordinate of ΨAn,+ is ±1. By Lemmas 4.8, 4.13, and 4.14.,we have ΨAk,+ is zero for k < n. This is equivalent to the lemma. 2
Lemma 4.16. The first row of m+ checks and the last row of m+ checks.
Proof. The first statement follows immediately from Lemma 4.14. The second statementfollows immediately from Lemma 4.11. 2
Now we finish the proof. Consider the ith row of m+. Let k = i + 2. In light of the trivialcases taken care of above, we can assume that 3 < k ≤ m. Let F = Ak,+. As we discussed inthe proof of Corollary 4.10, each polynomial ∂A/∂Fj has either 0 or 1 hefty terms.
Assume that j is even. Let J ⊂ 1, ..., 2n be the unoccupied stretch from Lemma 4.7. LetJ ′ ⊂ J denote the smaller set obtained by removing the first and last 3 members from J . It
33
follows from the construction in Lemma 4.7 that ∂F/∂j has a hefty term if and only if j ∈ J ′.Thus the jth entry of the kth row is ±1 if and only if j ∈ J ′. Similar considerations hold whenj is odd. It is an exercise to show that the conditions we have given translate precisely into theform given in (4.10). Hence m+ checks.
Remark 4.17. One can approach the proof differently. When we move from row k to row k+2the corresponding interval J ′ = (a, b) changes to the new interval J ′ = (a + 4, b − 4). Fromthis fact, and from our choice of minors, it follows easily that row k checks if and only if rowk + 2 checks. At the same time, when n is replaced by n+ 2, the interval J ′ = (a, b) changes toJ ′ = (a, b + 4). This translates into the statement that row k checks for n if and only if row kchecks for n + 2. All this reduces the whole problem to a computer calculation of the first fewcases. We did the calculation up to the case n = 13 and this suffices.
4.6 Completion of the second calculation
We make all the same definitions and conventions for the second calculation, using the matrix(family) in (4.14) in place of the matrix (family) in (4.10). The argument for the second calcula-tion is really just the same as the argument for the first calculation. Essentially, we just ignorethe outer 8 coordinates and see what we get. What makes this work is that all the functionsexcept An,± are great – the inner indices determine the heft. To handle the last row of m+,which involves the Casimir An,+, we use Lemma 4.12 in place of Lemma 4.11.
It remains to establish the Justification Lemma 4.4. It is convenient to define
δ =(n− 4)(n− 5)
2. (4.21)
We also mention several other pieces of notation and terminology. When we line up the indices5, ..., 2n − 4, there are 2 middle indices. When n = 7 the middle indices of 5, 6, 7, 8, 9, 10 are 7and 8. Let π⊥ denote the projection from R2n onto R8 obtained by stringing out the first andlast 4 coordinates.
Lemma 4.18 (Tangent Estimate). The following properties of π⊥(vj) hold:
• All coordinates are O(1).
• Coordinates 3 and 6 are O(u).
• Except when j is one of the middle two indices, coordinates 1 and 8 are O(uδ+1).
• When j is the first middle index, coordinate 1 is uδ+O(uδ+1) and coordinate 8 is O(uδ+1).
• When j is the second middle index, coordinate 8 is uδ + O(uδ+1) and coordinate 1 isO(uδ+1).
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Proof. We prove this in the next section. 2
Lemma 4.19. The Justification Lemma holds for F = An,+.
Proof. A direct calculation shows that, up to O(uδ+1),
Combining this last equation with our two lemmas, we have
(∇8F )j = ∇8F · ej = (Υ8F )j +O(u). (4.30)
This holds for all j. This completes the proof of the Justification Lemma.
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5 The polygon and its tangent space
The goal of this section is to construct the polygon P u and prove the Tangent Lemma, whichestimates the tangent space TPu(C). We will begin by repackaging some of the material workedout in [19]. The results here are self-contained, though our main formula relies on the work donein [19]. In order to remain consistent with the formulas in [19], we will use a slightly differentlabelling convention for polygons.
5.1 Polygonal rays
We say that a polygonal ray is an infinite list of points P−7, P−3, P1, P5, ... in the projectiveplane. We normalize so that (in homogeneous coordinates)
The first 4 points are normalized to be the vertices of the positive unit square, starting at theorigin, and going counterclockwise. Here we are interpreting these points in the usual affinepatch z = 1. This polygonal ray defines lines:
L−5+k = P−7+kP−3+k; k = 0, 4, 8, ... (5.2)
We denote by LL′ the intersection L ∩ L′. Similarly, PP ′ is the line containing P and P ′.The pairs of points and lines determine flags, as follows:
Remark 5.1. Notice that it is impossible to define χ(F−2) because we would need to knowabout a point P−11, which we have not suppled. Likewise, it is impossible to define χ(F−4)because we would need to know about L−9, which we have not supplied. Thus, the invariantsx0, x1, x2, ... are well defined for our polygonal ray.
37
Cross product in vector form: Since we are going to be computing a lot of these crossratios, we mention a formula that works quite well. We represent both points and lines inhomogeneous coordinates, so that (a, b, c) represents the line corresponding to the equationax+ bx+ cz = 0. We define V ∗W to be the coordinate-wise product of V and W . Of course,V ∗W is also a vector. Let (×) denote the cross product. We have
(χ, χ, χ) =(A×B) ∗ (C ×D)
(A× C) ∗ (B ×D). (5.7)
Here χ is the inverse cross ratio of the points or lines represented by these vectors. It mayhappen that some coordinates in the denominator vanish. In this case, one needs to interpretthis equation as a kind of limit of nearby perturbations. This formula works whenever A,B,C,Drepresent either collinear points or concurrent lines in the projective plane.
5.2 The reconstruction formulas
Referring to the definition of the monodromy invariants, we define Oba to be the sum over allodd admissible monomials in the variables x0, x1, x2, ... which do not involve any variables withindices i ≤ a or i ≥ b. For instance
O11 = 1, O3
1 = 1, O51 = 1− x3, O7
1 = 1− x3 + x3x4x5.
We also note that, when a < 0, the polynomial Oba is independent of the value of a. For thisreason, when a < 0 we simply write Ob in place of Oba. The corresponding set Sb consists ofadmissible sequences, all of terms are less than b.
Given a list (x0, x1, x2, ...) we seek a polygonal ray which has this list as its corner invariants.Here is the formula.
P9+2k = (O3+k −O3+k1 + x0x1O
3+k3 , O3+k, O3+k + x0x1O
3+k3 ), k = 0, 2, 4, ... (5.8)
We would also like a formula for reconstructing the lines of a polygonal ray. We start withthe obvious:
For the remaining points, we define polynomials Eba exactly as we defined Oba except we inter-change the uses of even and odd . Thus, for instance E6
2 = 1− x4. Here is the formula.
L7+2k = (E2+k − E2+k0 , E2+k
0 − x0E2+k2 ,−E2+k), k = 0, 2, 4... (5.10)
Remark 5.2. These formulas are equivalent to equations 19 and 20 in [19], but the normalizationof the first 4 points is different, and the roles of points and lines have been switched. We gotthe above formulas by applying a suitable projective duality to the polygonal ray in [19].
38
We mention one important connection between our various reconstruction formulas. Thefollowing is an immediate consequence of Lemma 3.2 in [19]:
We close this section with a characterization of the moduli space of closed polygons withinX. We do not need this result for our proofs, but it is nice to know.3
Lemma 5.3. The invariant x1, ..., x2n define a closed polygon if and only if O2n−5 and all itscyclic shifts vanish.
Proof. We can think of a closed polygon as an n-periodic infinite ray. The periodicity impliesthat P4n−7 = P−7 = (0, 0, 1). Since 4n− 7 = 2k + 9 for k = 2n− 8, equation (5.8) tells us thatO2n−5 = 0. Considering equation 5.10, we see that E2n−6 = E2n−6
0 = 0. But E2n−60 is a cyclic
shift of O2n−5. Hence, if P is closed then O2n−5 and all its cyclic shifts vanish.Conversely, if O2n−5 and all its shifts vanish then P4n−7 ∈ L−5 and P−3 ∈ L4n−5. Likewise
P4n−3 ∈ L−1 and P1 ∈ L4n−1, and so on. This situation forces P4n−3 = P−3. Shifting theindices, we see that P4n+1 = P1, and so on. 2
Remark 5.4. Observe that O2n−5 involves exactly 2n− 7 consecutive corner invariants. If thefirst 2n − 8 are specified, then the next variable can be found by solving O2n−5 = 0. Thus,Lemma 5.3 gives an algorithmic way to find a closed n-gon whose first 2n− 8 corner invariantsare specified.
5.3 The polygon
We start with an infinite periodic list of variables which starts out
(u, u2, ..., un−4, un−4, ..., u2, u1, ...) (5.12)
and has period 2n− 8. We let Xu denote the polygonal ray associated to this infinite list. Onceu is sufficiently small, the first n points of Xu are well defined. We define P u to be the n-gonmade from the first n-points of Xu, and we take u small enough so that this definition makessense.
The first 2n − 8 corner invariants of P u, which we now identify with x0, ..., x2n−9, are theones listed in equation (5.12). However, when it comes time to compute x2n−8, ..., x2n−1, we donot use the relevant points of Xu but rather substitute in the corresponding point of P u. Thus,the remaining 8 corner invariants change. We write the corner invariants of P u as
a, b, c, d, u, u2, u3, ..., u3, u2, u, d′, c′, b′, a′. (5.13)
3One could give an alternative proof of Proposition 3.1 computing the Poisson bracket of the polynomials ofLemma 5.3 with the monodromy invariants.
39
It follows from symmetry that e = e′ for each e ∈ a, b, c, d. This symmetry here is that thefirst 2n − 8 invariants determine P , and their palindromic nature forces P to be self-dual: theprojective duality carries P to the dual polygon made from the lines extending the sides of P .
Lemma 5.5. e = 1 +O(u) for each e ∈ b, c, d.
Proof. We set P−11 = (X,Y, Z) and L−13 = (U, V,W ). We have
Looking at the signs in equation (5.21), we see that a = us +O(us+1). 2
5.4 The tangent space
Recall that π : R2n → R2n−8 is the projection which strips off the outer 4 coordinates. Let π⊥
be as in the Tangent Estimate Lemma 4.18. Recall that vk is the special basis of TP (C) suchthat π(vk) = ek for k = 5, ..., 2n− 4.
Lemma 5.7. The following holds concerning the coordinates of π⊥(vj):
• Coordinates 2, 4, 5, 7 of π⊥(vj) have size O(1)
• Coordinates 3, 6 have size O(u).
Proof. As above, we will just consider coordinates 2, 3, 4. The other cases follow from sym-metry.
We refer to the quantities used in the proof of Lemma 5.5. Each of these quantities is apolynomial in the coordinates, depending only on n. Hence dX/dt, etc., are all of size at mostO(1). Moreover, the denominators on the right hand sides of equations (5.16), (5.17), and(5.18) are all O(1) in size. Our first claim now follows from the product and quotient rules ofdifferentiation.
For our second claim, we differentiate equation (5.17):
dc
dt=X ′(Y − Z)−X(Y ′ − Z ′) + ZY ′ − Y Z ′
(X + Z)2=∗
Y ′ − Z ′ +O(u) =d
dt(−x0x1)O3+k
3 . (5.25)
The starred equality comes from the fact that (X,Y, Z) = (0, 1, 1)+O(u). The claim now followsfrom the fact that x0(0) = u and x1(0) = u2 and (O3+k
3 )′(0) = O(1). 2
41
Lemma 5.8. The following holds concerning the coordinates of π⊥(vj):
• When j is not a middle index, coordinates 1 and 8 of are of size O(uδ+1).
• When j is the first middle index, coordinate 1 equals uδ(1 + O(u)) and coordinate 8 is ofsize O(uδ+1).
• When j is the second middle index, coordinate 8 equals uδ(1 + O(u)) and coordinate 1 isof size O(uδ+1).
Proof. We will just deal with coordinate 1. The statements about coordinate 8 follow fromsymmetry.
Let us revisit the proof of Lemma 5.6. Let f = −(A×B)2. We have a = fg, where
g = − (C ×D)2(A× C)2(B ×D)2
. (5.26)
We imagine that we have taken some variation, and all these quantities depend on t.Each of the factors in the equation for g has derivative of size O(1). Moreover, the denomi-
nator in g has size O(1). From this, we conclude that
Using the variablesx1 = u, ..., xj = uj + t, xj+1 = uj+1, ... (5.32)
we get the result of this lemma as a simple exercise in calculus. 2
42
The results above combine to prove the Tangent Space Lemma.
Acknowledgments. Some of this research was carried out in May, 2011, when all three authorswere together at Brown University. We would like to thank Brown for its hospitality duringthis period. VO was partially supported by the PICS “PENTAFRIZ” of CNRS. RES waspartially supported by N.S.F. Grant DMS-0072607, and by the Brown University Chancellor’sProfessorship. ST was partially supported by the Simons Foundation grant No 209361 and bythe NSF grant DMS-1105442. We are grateful to the referee for numerous useful suggestions.
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